Two parallel plates of area $A$ are separated by two different dielectrics as shown in the figure. Find the resultant capacitance.

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance:

The parallel combination of two air-filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage $V$. When the capacitors are fully charged,the battery is removed,and after that,a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is completely filled with a dielectric material of constant $K$,then the potential difference across the other capacitor will become:

When a parallel plate capacitor is charged up to $95 \ V$,its capacitance is $C$. If a dielectric slab of thickness $2 \ mm$ is inserted between the plates and the distance between the plates is increased by $1.6 \ mm$ such that the same potential difference is maintained,the dielectric constant of the material (slab) is:

The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants \(K _1=2, K_2=3\), and \(K _3=5\) (as shown in the figure). If new capacitance is \(\frac{ n }{3} C\) then the value of \(n\) is ___________.